codeforces 507A Amr and Music_k - amr and music codeforces

本文链接：https://blog.youkuaiyun.com/zhang__liuchen/article/details/76726553

本文介绍了一个关于如何合理分配有限的学习时间来掌握多种乐器的问题，并提供了一段C++代码实现。问题的核心在于选择能在限定时间内学会的最大数量的乐器。

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Amr is a young coder who likes music a lot. He always wanted to learn how to play music but he was busy coding so he got an idea.

Amr has n instruments, it takes ai days to learn i-th instrument. Being busy, Amr dedicated k days to learn how to play the maximum possible number of instruments.

Amr asked for your help to distribute his free days between instruments so that he can achieve his goal.

Input

The first line contains two numbers n, k (1 ≤ n ≤ 100, 0 ≤ k ≤ 10 000), the number of instruments and number of days respectively.

The second line contains n integers ai (1 ≤ ai ≤ 100), representing number of days required to learn the i-th instrument.

Output

In the first line output one integer m representing the maximum number of instruments Amr can learn.

In the second line output m space-separated integers: the indices of instruments to be learnt. You may output indices in any order.

if there are multiple optimal solutions output any. It is not necessary to use all days for studying.

   Examples
 
     input
   
4 10
4 3 1 2

     output
   
4
1 2 3 4

     input
   
5 6
4 3 1 1 2

     output
   
3
1 3 4

     input
   
1 3
4

     output
   
0

Note

In the first test Amr can learn all 4 instruments.

In the second test other possible solutions are: {2, 3, 5} or {3, 4, 5}.

In the third test Amr doesn't have enough time to learn the only presented instrument.

题意懂了就会了。

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
struct node
{
    int x,y;
} a[110];
int comp(node x,node y)
{
    return x.x<y.x;
}
int main()
{
    int n,k,b[110],w=0;
    scanf("%d%d",&n,&k);
    for(int i=0; i<n; i++)
    {
        scanf("%d",&a[i].x);
        a[i].y=i;
    }
    sort(a,a+n,comp);
    for(int i=0;i<n;i++)
    {
        if(k>=a[i].x)
        {
            k-=a[i].x;
            b[w++]=a[i].y;
        }
    }
    printf("%d\n",w);
    for(int j=0;j<w;j++)
    {
        if(j==0)
            printf("%d",b[j]+1);
        else
            printf(" %d",b[j]+1);
    }
    if(w)
        printf("\n");
    return 0;
}